In the figure, ∠BAD = 65° , ∠ABD = 70° , ∠BDC = 45°
(i) Prove that AC is a diameter of the circle
(ii) Find ∠ACB
(i) Prove that AC is a diameter of the circle
(ii) Find ∠ACB
Exercise 17 (A) | Q 2 | Page 257
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(i)
In Δ ABD ,
∠ DAB + ∠ ABD + ∠ ADB = 180°
⇒ 65° + 70° + ∠ ADB = 180°
⇒ 135° + ∠ ADB = 180°
⇒ ∠ ADB = 180° - 135° = 45°
Now , ∠ ADC = ∠ ADB + ∠ BDC = 45° + 45° = 90°
Since ∠ ADC is the angle of semicircle , so AC is a diameter of the circle.
(ii)
∠ ACB = ∠ ADB .....(angles in the same segment of a circle)
⇒ ∠ ACB = 45°
In Δ ABD ,
∠ DAB + ∠ ABD + ∠ ADB = 180°
⇒ 65° + 70° + ∠ ADB = 180°
⇒ 135° + ∠ ADB = 180°
⇒ ∠ ADB = 180° - 135° = 45°
Now , ∠ ADC = ∠ ADB + ∠ BDC = 45° + 45° = 90°
Since ∠ ADC is the angle of semicircle , so AC is a diameter of the circle.
(ii)
∠ ACB = ∠ ADB .....(angles in the same segment of a circle)
⇒ ∠ ACB = 45°
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